Posted by **jackie** on Tuesday, December 1, 2009 at 4:47pm.

f(x)= x^2 / (x-2)^2

find the intervals of concavity and the inflections points

- calc -
**bobpursley**, Tuesday, December 1, 2009 at 4:48pm
I will be happy to critique your work.

- calc -
**MathMate**, Tuesday, December 1, 2009 at 5:11pm
If the function is twice differentiable within its domain, inflection points occur at values of c where f"(c)=0.

Note that not all points where f"(c)=0 are inflection points.

You need to check the concavity *between* points where f"(x)=0. It is concave up if f"(x)>0 and concave down if f"(x)<0.

X=c is an inflection point if all of the following conditions are satisfied:

1. f(x) is twice differentiable at c.

2. f"(c)=0

3. f"(c+) has a different sign than f"(c-), i.e. concavity changes.

Now it's time for you to sharpen your pencil and show us some work.

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